Question

The energy levels of an atom of a certain element are shown in the given figure 5. Which one of the transitions A, B, C, D or E will result in the emission of photons of electromagnetic radiation of wavelength 618.75 nm?
Support your answer with mathematical calculations.

Answer 1

$\Delta E=\frac{hc}{\lambda }$
$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{618.75\times {10}^{-9}}=3.2\times {10}^{-19}J=2eV$
This corresponds to the transition D for which
$\Delta E=(-1)-(-3eV)=2eV$